# Homework Help: Problems with Electric Field questions

1. Mar 13, 2008

### User1247

1. The problem statement, all variables and given/known data
Find the point on the x-axis where the net electric field is zero for two particles of charges q1=1x10^-9C and q2=2x10^-9C. Assume q1 and q2 are 20cm apart. Assume q1 is located at x=0.

I solved this but it's not the same as the book's answer.. GAH! What do you guys get? Thanks.

2. Relevant equations
Enet=E1+E2, Enet=0 => |E1|=|E2|

E=kq/r² where q is the charge creating the field

3. The attempt at a solution
E1=kq1/r²
E2=kq2/(.2m - r)²

kq1/r²=kq2/(.2m - r)²
q1/r²=q2/(.2m - r)²
q1(.2m - r)²=q2r²
q1(.2m - r)²=q2r²
q1(0.4m-0.4mr+r²)-q2r²=0 <------ There's my mistake
(q1)0.4m-(q1)0.4mr+(q1)r²-q2r²=0
(q1-q2)r²-(q1)0.4mr+(q1)0.4m=0
((1E-9C)-(2E-9C))r²-(1E-9C)0.4mr+(1E-9C)0.4m=0
(-1E-9C)r²-(1E-9C)0.4mr+(1E-9C)0.4m=0
(-1E-9C)r²-(4E-10C)mr+(4E-10C)m=0
(-1E-9)r²-(4E-10)r+(4E-10)=0
r=-0.863325 or r=0.463325

The answer in the book is 8.3cm. Where did I go wrong?
------
Edit:
q1(.2m - r)²=q2r²
q1(0.04m-0.4mr+r²)-q2r²=0
(q1)0.04m-(q1)0.4mr+(q1)r²-q2r²=0
(q1-q2)r²-(q1)0.4mr+(q1)0.04m=0
((1E-9C)-(2E-9C))r²-(1E-9C)0.4mr+(1E-9C)0.04m=0
(-1E-9C)r²-(1E-9C)0.4mr+(1E-9C)0.04m=0
(-1E-9C)r²-(4E-10C)mr+(4E-11C)m=0
(-1E-9)r²-(4E-10)r+(4E-11)=0
Positive r comes out to 8.3cm

Last edited: Mar 13, 2008
2. Mar 13, 2008

### Oerg

do you realise that Q2 has charge two times of Q1? Simply suibstitute Q2 with 2Q1 and cancel out the Q1s and your equations will appear much simpler.

I managed to simplify the equation into a quadratic equation offhand. You should arrive at a quadratic equation.

Last edited: Mar 13, 2008
3. Mar 13, 2008

### User1247

Right, you can see:

(-1E-9)r²-(4E-10)r+(4E-10)=0 before I found my mistake and
(-1E-9)r²-(4E-10)r+(4E-11)=0 after I found my mistake.